= Solution
For the value condition, a <function> agrees with itself outside the empty exceptional <set>, proving the <reflexive relation> property. Interchanging the two <functions> preserves equality, so the same exceptional <finite set> proves the <symmetric relation> property.
Suppose $(A,f)$ is related to $(B,g)$ and $(B,g)$ to $(C,h)$, with exceptional <finite sets> $F_{AB}$ and $F_{BC}$. Points of $A\cap C$ might be absent from $B$, so merely taking $F_{AB}\cup F_{BC}$ would miss a real issue. Use instead
$$
F_{AC}=(A\cap C)\cap\bigl(F_{AB}\cup F_{BC}\cup(A\setminus B)\bigr).
$$
This is a <finite set> contained in $A\cap C$: the first two terms are finite by hypothesis and $A\setminus B\subseteq A\mathbin\triangle B$ is finite by the domain condition. If $x\in(A\cap C)\setminus F_{AC}$, then $x\in B$ and neither original exception occurs, so $f(x)=g(x)=h(x)$. The domain condition is also preserved, as proved in the preceding scope. Thus the full relation is a <transitive relation>, and
$$
\boxed{\mathcal R\text{ is an equivalence relation on }\Sigma.}
$$
Finite domains cause no problem: one may take the whole <intersection> as the exceptional <set>. Both conditions are needed for <equivalence of partial functions modulo finite changes>.
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